Solve yz^2+(4+3i)z-12-11i=0

$$y z ^ { 2 } + ( 4 + 3 i ) z - 12 - 11 i = 0$$

Solve for y

$$y=-\frac{\left(4+3i\right)z+\left(-12-11i\right)}{z^{2}}$$

$$z\neq 0$$

Steps for Solving Linear Equation

Subtract \left(4+3i\right)z from both sides. Anything subtracted from zero gives its negation.

$$yz^{2}-12-11i=\left(-4-3i\right)z$$

Add 12 to both sides.

$$yz^{2}-11i=\left(-4-3i\right)z+12$$

Add 11i to both sides.

$$yz^{2}=\left(-4-3i\right)z+12+11i$$

The equation is in standard form.

$$z^{2}y=\left(-4-3i\right)z+\left(12+11i\right)$$

Divide both sides by z^{2}.

$$\frac{z^{2}y}{z^{2}}=\frac{\left(-4-3i\right)z+\left(12+11i\right)}{z^{2}}$$

Dividing by z^{2} undoes the multiplication by z^{2}.

$$y=\frac{\left(-4-3i\right)z+\left(12+11i\right)}{z^{2}}$$

Solve for z

$$\left\{\begin{matrix}z=\frac{\sqrt{\left(48+44i\right)y+\left(7+24i\right)}+\left(-4-3i\right)}{2y}\text{; }z=\frac{-\sqrt{\left(48+44i\right)y+\left(7+24i\right)}+\left(-4-3i\right)}{2y}\text{, }&y\neq 0\\z=\frac{81}{25}+\frac{8}{25}i\text{, }&y=0\end{matrix}\right.$$